# Q Gained = Q Lost

1. Jan 21, 2009

### Soliloquy12

1. The problem statement, all variables and given/known data

Solve algebraically for T2 then find T2.

T1C=63 Celsius
T1H=79.5 Celsius
MCW=8.63g
MBW=36.48g

2. Relevant equations

QLost = QGained

-mbwcw(T2-T1H) = mcwcw(T2-T1C)

3. The attempt at a solution

Trying to solve algebraically I arrived at:

T2 = mcwt1C+mbwt1h$$/$$(mbw+mcw)

But when substituting the values I get T2 = 123.66J from the original equation and T2 = 76.34 when using the equation I derived from the original.

Can someone please show me where I am going wrong with the equation here?

Thanks!

2. Jan 21, 2009

### Staff: Mentor

What do you mean? How exactly did you get that result? (I assume that's the final temperature of some mixture, not an energy in Joules.)
I didn't do the calculation, but if the problem is what I think it is, that sounds reasonable.

3. Jan 21, 2009

### Soliloquy12

Yes sorry i meant degrees celsius. I substitutted the values into the initial "long" equation when I got 123.66 degrees celsius.

4. Jan 21, 2009

### Staff: Mentor

I still don't know what you mean. The initial "long" equation is the same equation that you rearranged to solve for T2. All you did was solve it algebraically--it's still the same equation.

Show exactly what you did. What values did you substitute?

5. Jan 21, 2009

### AEM

First, I would suggest that you put parentheses around the two terms in your numerator for clarity and to prevent mistakes.

Second, I'm assuming that this problem involves adding something hot to something cool because you didn't really say. In problems such as that the final temperature will lie between the two starting temperatures. Therefore you should suspect that you made a simple algebraic error when you computed 123.66 degrees. As Dr AL indicated, without seeing the details of your calculation, one can't say more.